An olympic diver springs forward off a 10 meter platform. He enters the water 8 meters beyond the edge of the platform. how much time does it spend in the air?

The key is to assume that he takes off horizontally. Few Olympic divers do that. You are in the air less and can't do as many maneuvers.

The time to hit the waterl is in this case
t = sqrt(2H/g) = 1.429 seconds

Where he lands doesn't matter. That depends upon his horizontal takeoff velocity

9.8

Q13

To find the time the diver spends in the air, we can use the equation for motion under constant acceleration. In this case, the acceleration is due to gravity (approximately 9.8 m/s^2). We know the initial velocity (0 m/s) since the diver springs forward, and we need to find the time (t) it takes for the diver to cover the horizontal distance of 8 meters.

The formula we'll use is:

d = v0 * t + (1/2) * a * t^2

Where:
d is the distance covered
v0 is the initial velocity
a is the acceleration
t is the time

Since the horizontal distance covered is 8 meters, the initial velocity is 0 m/s (as mentioned earlier), and the acceleration is -9.8 m/s^2 (negative because it's in the opposite direction of the motion), we can rearrange the equation to solve for time (t):

8 = 0 * t + (1/2) * (-9.8) * t^2
8 = (1/2) * (-9.8) * t^2
t^2 = 8 * 2 / (-9.8)
t^2 = 16 / (-9.8)
t^2 = -1.6326
t ≈ √(-1.6326) (taking the square root of both sides, considering only the positive value)

However, the result is imaginary, which means the equation has no real solutions. It implies that the diver does not spend any time in the air.

Therefore, the time the diver spends in the air is effectively 0 seconds.